The area of the quadrilateral whose sides are given by $\left|x\right|+\left|y\right|=1$  is 

Given sides of a quadrilateral $\left|x\right|+\left|y\right|=1$

$\therefore x+y=1,x-y=1,-x+y=1,-x-y=1$

By the graph given quadrilateral is a square of side $\sqrt{2}$

Since $\Delta OAB$ is right angle and $O{A}^2+O{B}^2=A{B}^2$

$1^2+1^2=A{B}^2$

$\therefore AB=\sqrt{2}$

$\therefore$ Area $\left(A\right)\text{\hspace{0.17em}\hspace{0.17em}is}=s^2=A{B}^2={\left(\sqrt{2}\right)}^2=2$ units